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Math 106 Exam 2 Solutions 15 December, 2005
- Consider f (x) = x^2 x^2 โ 1
The domain of f is R \ {โ 1 , 1 }. Since lim xโ 1 โ^ f (x) = โโ, lim xโ 1 +^ f (x) = +โ, lim xโโ 1 โ^ f (x) = +โ, and lim xโโ 1 +^ f (x) = โโ,
x = 1 and x = โ1 are vertical asymptotes.
Since (^) xโlim+โ f (x) = 1 and (^) xโโโlim f (x) = 1, y = 1 is the horizontal asymptote.
f โฒ(x) = 2 x(x^2 โ 1) โ x^2 (2x) (x^2 โ 1)^2
2 x^3 โ 2 x โ 2 x^3 (x^2 โ 1)^2
โ 2 x (x^2 โ 1)^2
So f โฒ(x) = 0 at x = 0 and undefined at x = 1 and x = โ1.
f โฒโฒ(x) = โ2(x^2 โ 1)^2 + 2x2(x^2 โ 1)2x (x^2 โ 1)^4
2(x^2 โ 1)(โx^2 + 1 + 4x^2 ) (x^2 โ 1)^4
2(3x^2 + 1) (x^2 โ 1)^3
So f โฒโฒ(x) is undefined at x = 1 and x = โ1. Then
x โ 1 0 1 f โฒ^ + + + || + + + 0 โ โ โ || โ โ โ f โฒโฒ^ + + + || โ โ โ โ โ โ || + + + f โ || โ 0 โ || โ So we get a local maximum value f (0) = 0 at x = 0. There are no local minimum and inflection points. Combining all this information we get the following graph.
- Let f (x) be defined piecewise as follows:
f (x) =
x^2 if 0 โค x < 1 โx + 2 if 1 โค x โค 2
a) We will find
0
f (x)dx using upper Riemann sums with subintervals of equal length. But
as the region can be divided into two subregions as in the above figure, we will find the upper Riemann sums for both of these subregions. Then the first upper Riemann sum, say S 1 is:
S 1 =
n f (
n
n f (
n
n f ( n n
n
[
n^2
n^2
n^2 n^2
]
So S 1 =
n^3
โ^ n
k=
k^2 =
n^3
n(n + 1)(2n + 1) 6
2 n^2 + 3n + 1 6 n^2
Next the second upper sum is :
S 2 =
n f (1) +
n f (1 +
n
n f (1 +
n
n f (1 + n โ 1 n
n
[
n
n
n โ 1 n
)]
n
[
n
n
n โ 1 n
)]
So S 2 =
n
[
n โ
n
nโโ 1
k=
k
]
n
[
n โ
n
(n โ 1)n 2
]
n โ 1 2 n
n + 1 2 n
Then
0
f (x)dx = (^) nโlim+โ(S 1 + S 2 ) = (^) nโlim+โ
2 n^2 + 3n + 1 6 n^2
n + 1 2 n
b)
0
f (x)dx =
0
x^2 dx +
1
(โx + 2)dx =
[
x^3 3
] 1
0
[
โx^2 2
] 2
1
- a) The average value of f (x) on [0, 2] is: 1 2
0
(2x + 1)^3
dx = 25
0
2(2x + 1)โ^3 dx = 25
[
(2x + 1)โ^2 โ 2
] 2
0
b) d dx
3 x^2 + 1 +
โซ (^3) x (^2) +
1
sin u u du
= 6x + d dx
โซ (^3) x (^2) +
1
sin u u du = 6x + sin(3x^2 + 1) 3 x^2 + 1 (6x).
c) Let A be the area of the region enclosed by y = x/2, y =
8 โ x and the x-axis. If we calculate with respect to y we obtain:
A =
0
(8 โ y^2 โ 2 y)dy =
[
8 y โ y^3 3 โ y^2
] 2
0
On the other hand if we calculate with respect to x we obtain:
A =
0
x 2
dx +
4
(8 โ x)^1 /^2 dx =
[
x^2 4
] 4
0
[
(8 โ x)^3 /^2 3 / 2
] 8
4
Extra credit question.
Let f (x) be a function such that f โฒโฒ(x) exists for all x. Suppose that the equation f (x) = x has at least three solutions. Then consider g(x) = f (x) โ x with three roots, say x 1 < x 2 < x 3. Since g(x) is continuous and differentiable, by the Mean Value Theorem there
is a point c 1 in (x 1 , x 2 ), where gโฒ(c 1 ) = g(x 2 ) โ g(x 1 ) x 2 โ x 1 = 0. Similarly by the Mean Value
Theorem there is a point c 2 in (x 2 , x 3 ), where gโฒ(c 2 ) = g(x 3 ) โ g(x 2 ) x 3 โ x 2 = 0. But since gโฒ(x) is
also continuous and differentiable, if we apply the Mean Value Theorem again we get there
exists a point c in (c 1 , c 2 ) with f โฒโฒ(c) = gโฒโฒ(c) = gโฒ(c 2 ) โ gโฒ(c 1 ) c 2 โ c 1
